How to derive the expression for cosh^(-1)z = log[z + (z^2-1)^(1/2)]?

Question

I have to prove that

cosh^(-1) z = log[z + (z^2-1)^(1/2)]

I just want to know what formula do I start with to get to the above.

I know that cosh(y) = [e^y + e^(-y)]/2

So if that is the equation I should use... how do i inverse it ... I read somewhere that you just switch y's and x's... so it would be y = [e^x + e^(-x)] / 2
2y = e^x + e^(-x)

I'm not good enough working with logs to even continue, i feel like it is probably the wrong start anyway... can anyone please let me know where to start... thanks :)

Laura A (working at midnight on maths again as usual!)

Answer 1

First, you need to note that

cosh(ln(z + (z² – 1)½)) =
(e^( ln(z + (z² – 1)^½)) + e^(- ln(z + (z² – 1)^½)))/2
= (z + (z² – 1)^½ + e^(ln(1/(z + (z² – 1)^½))))/2
= (z + (z² – 1)^½ + 1/(z + (z² – 1)^½))/2
= (z + (z² – 1)^½ +
(z – (z² – 1)^½) /((z + (z² – 1)^½) (z – (z² – 1)^½)))/2
= (z + (z² – 1)^½ + (z – (z² – 1)^½) / (z² – (z² – 1)))/2
= (z + (z² – 1)^½ + z – (z² – 1)^½)/2
= 2z /2
= z

then

z = cosh(ln(z + (z² – 1)^½))

Now cosh^(-1) is a function, such that, cosh^(-1)(cosh(x)) = x

Thus

cosh^(-1)z = ln(z + (z² – 1)^½)

Any question please don’t doubt to contact me